README_EN.md

16. 3Sum Closest

中文文档

Description

Given an integer array nums of length n and an integer target, find three integers in nums such that the sum is closest to target.

Return the sum of the three integers.

You may assume that each input would have exactly one solution.

 

Example 1:

Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

Example 2:

Input: nums = [0,0,0], target = 1
Output: 0
Explanation: The sum that is closest to the target is 0. (0 + 0 + 0 = 0).

 

Constraints:

• 3 <= nums.length <= 500
• -1000 <= nums[i] <= 1000
• -104 <= target <= 104

Solutions

Python3

class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        def twoSumClosest(nums, start, end, target):
            res = 0
            diff = 10000
            while start < end:
                val = nums[start] + nums[end]
                if val == target:
                    return val
                if abs(val - target) < diff:
                    res = val
                    diff = abs(val - target)
                if val < target:
                    start += 1
                else:
                    end -= 1
            return res

        nums.sort()
        res, n = 0, len(nums)
        diff = 10000
        for i in range(n - 2):
            t = twoSumClosest(nums, i + 1, n - 1, target - nums[i])
            if abs(nums[i] + t - target) < diff:
                res = nums[i] + t
                diff = abs(nums[i] + t - target)
        return res

Java

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int res = 0;
        int n = nums.length;
        int diff = Integer.MAX_VALUE;
        for (int i = 0; i < n - 2; ++i) {
            int t = twoSumClosest(nums, i + 1, n - 1, target - nums[i]);
            if (Math.abs(nums[i] + t - target) < diff) {
                res = nums[i] + t;
                diff = Math.abs(nums[i] + t - target);
            }
        }
        return res;
    }

    private int twoSumClosest(int[] nums, int start, int end, int target) {
        int res = 0;
        int diff = Integer.MAX_VALUE;
        while (start < end) {
            int val = nums[start] + nums[end];
            if (val == target) {
                return val;
            }
            if (Math.abs(val - target) < diff) {
                res = val;
                diff = Math.abs(val - target);
            }
            if (val < target) {
                ++start;
            } else {
                --end;
            }
        }
        return res;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var threeSumClosest = function (nums, target) {
    let len = nums.length;
    nums.sort((a, b) => a - b);
    let diff = Infinity;
    let res;
    for (let i = 0; i < len - 2; i++) {
        if (i > 0 && nums[i] === nums[i - 1]) continue;
        let left = i + 1,
            right = len - 1;
        let cur = nums[i] + nums[i + 1] + nums[i + 2];
        if (cur > target) {
            let newDiff = Math.abs(cur - target);
            if (newDiff < diff) {
                diff = newDiff;
                res = cur;
            }
            break;
        }
        while (left < right) {
            cur = nums[i] + nums[left] + nums[right];
            if (cur === target) return target;
            let newDiff = Math.abs(cur - target);
            if (newDiff < diff) {
                diff = newDiff;
                res = cur;
            }
            if (cur < target) {
                while (nums[left] === nums[left + 1]) left++;
                left++;
                continue;
            } else {
                while (nums[right] === nums[right - 1]) right--;
                right--;
                continue;
            }
        }
    }
    return res;
};

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